Let $M$ be a manifold and $A$ a $*$-algebra. Does is hold that $$C^{\infty}(M,A) \cong C^{\infty}(M) \otimes A$$ where the RHS means that you take smooth functions which map into $A$. If this holds, do you have a proof?
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4$\begingroup$ If $A$ is finite dimensional, this is trivial (pick a basis for $A$). If not, then you need to be more specific on the conditions you agree to on $A$ (topology, etc.). $\endgroup$– Igor KhavkineCommented Jun 3, 2014 at 11:12
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6$\begingroup$ Since $C^\infty(M)$ is a nuclear Frechet space you can choose more or less any of the usual tensor topologies (like the projective $\pi$ or the injective $\varepsilon$) and you get $C^\infty(M,A)\cong C^\infty(M) \tilde{\otimes} A$ (the completed tensor product) for all Banach (or Frechet) spaces $A$. $\endgroup$– Jochen WengenrothCommented Jun 3, 2014 at 12:01
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